Standing sound waves created in a pipe, question about resonances

AI Thread Summary
In a closed pipe, resonance occurs at specific intervals of the wavelength, which are determined by the formula fn = n(v/4L). For a pipe length of approximately one meter and sound velocity of 340 m/s, the wavelength for 500 Hz is calculated, revealing three resonance cases. Similarly, for 700 Hz, four resonance cases are identified, and for 1000 Hz, six cases arise. The number of resonance cases corresponds to how many quarter wavelengths fit within the length of the pipe. Understanding these principles clarifies the relationship between frequency, wavelength, and resonance in a closed pipe system.
helpmeplz!
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Hey guys i have a question for my lab report that i can't solve. We did a lab where we had a pipe and produced sound resonance in there, using a generator to vary the frequencies and we recorded our data.

the general formula is for the closed pipe, fn= n (v/4L) and wavelenght= 4L/n.

The question is, explain why there are 3 cases of resonance when f= 500 hz, 4 cases of resonance when f= 700 hz and 6 cases of resonance when f= 1000 hz.

the length of the pipe was around 1 metre.
 
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Hi helpmeplease!, welcome to PF.
In a closed pipe, resonance occurs when the length of pipe is λ/4, 3λ/4, 5λ/4, and so on.
Length of the pipe is nearly one meter. If you take the velocity of the sound at temperature as 340 m/s, what is the wavelength of 500 Hz sound? Then see how many λ/4 lengths fit in one meter.
 

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